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Kruka [31]
2 years ago
10

A young kid of mass m = 36 kg is swinging on a swing. The length from the top of the swing set to the seat is L = 3.5 m. The boy

is attempting to swing all the way around in a full circle. What is the minimum speed, in meters per second, the boy must be moving with at the top of the path in order to make a full circle? v = Assuming the boy is traveling at the speed found in part (a), what is their apparent weight. Wa in newtons, at the top of their path? (At the top. the boy is upside-down.) What is the boy's apparent weight, in newtons, at the bottom of their path if they have the velocity from part (a) at the top?
Physics
1 answer:
lara31 [8.8K]2 years ago
6 0

Answer:

Explanation:

Given

mass of boy=36 kg

length of swing=3.5 m

Let T be the tension in the swing

At top point mg-T=\frac{mv^2}{r}

where v=velocity needed to complete circular path

Th-resold velocity is given by mg-0=\frac{mv^2}{r}

v=\sqrt{gr}=\sqrt{9.8\times 3.5}=5.85 m/s

So apparent weight of boy will be zero at top when it travels with a velocity of v=\sqrt{gr}

To get the velocity at bottom conserve energy at Top and bottom

At top E_T=mg\times 2L+\frac{mv^2}{2}

Energy at Bottom E_b=\frac{mv_0^2}{2}

Comparing two as energy is conserved

v_0^2=4gl+gl

v_0^2=5gL

v_0=\sqrt{5gL}=13.09 m/s

Apparent weight at bottom is given by

W=\frac{mv_0^2}{L}-mg=\frac{36\times 13.09^2}{3.5}+36\times 9.8=2115.23 N

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A gymnast of mass 63.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
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B )

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= 618.03 N

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T =  mg + m a

= m ( g + a )

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D )  If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2

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A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at
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Answer:

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

Explanation:

Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K

mg=k x

1 x 32= k x 2

K=16

And also give that damping force is 8 times the velocity so damping constant C=8.

We know that equation for spring mass system

my''+Cy'+Ky=F

Now by putting the values

1 y"+8 y'+ 16y=6 cos 4 t ----(1)

The general solution of equation Y=CF+IP

Lets assume that at steady state the equation of y will be

y(IP)=A cos 4t+ B sin 4t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -4A sin 4t+4B cos 4t

y"=-16A cos 4t-16B sin 4t

Now put the values of y" , y' and y in equation 1

1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

So by comparing the coefficient both sides

-16A+32B+16A=0  So B=0

-16 B-32 A+16B=6  So A=-3/16

y=-3/16 cos 4t

Now to find the CF  of differential equation 1

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Homogeneous version of above equation

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So the general equation

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C_1=\dfrac{3}{16}

t=0 Y'=0 So

C_2 =\dfrac{3}{8}

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

The above equation is the general equation for motion.

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